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What is the median of set \(S =\{a  b, b  a, a + b\}\) ? (1) The mean of set \(S\) is equal to \(a + b\). (2) The range of set \(S\) is equal to \(2b\).
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16 Sep 2014, 01:10
Official Solution: (1) The mean of set \(S\) is equal to \(a + b\). Given that \(mean=\frac{(a  b)+(b  a)+(a + b)}{3}=a+b\), which leads to \(a+b=0\). Now, if \(a+b=0\), then \(ab\) and \(ba\) are either both zeros (if \(a=b=0\)) or have different signs (if \(a \ne b\)). In any case the median of \(S\) is \(a+b=0\). Sufficient. (2) The range of set \(S\) is equal to \(2b\). If \(a=b=0\), then \(median=0\) but if \(a=0\) and \(b=1\), then \(median=1\). Not sufficient. Answer: A
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Re: M2102
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07 Sep 2016, 02:05
Manmeet0 wrote: Bunuel wrote: manhasnoname wrote: Option (1) states mean is a+b, which can be inferred from the question stem.
How does this lead a+b to 0? Could someone please clarify? By simplifying this: \(mean=\frac{(a  b)+(b  a)+(a + b)}{3}=a+b\) doesnt simlifying leads to: a+B/3=a+b ???????????? Yes, but we can simplify further: (a + b)/3 = a + b; a + b = 3(a + b); 2(a + b) = 0; a + b = 0.
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Re: M2102
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01 Jan 2017, 08:18
arven wrote: I worked the second statement as following:
Range= Highest Number  Lowest Number R=2b In order for R to be 2b, the only possible combination is (b+a)(ab). Thus, (ab)≤(ba)<(b+a) or (ab)<(ba)≤(b+a) So ba is the median.
Where am I wrong? (2) The range of set \(S\) is equal to \(2b\). If \(a=b=0\), then \(median=0\) but if \(a=0\) and \(b=1\), then \(median=1\). Not sufficient.
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Re: M2102
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04 Oct 2014, 06:26
In option 2, The range (largest  Least )= 2b; how would be arrive at 2b from the set of the three numbers viz : a+b, ba, ab ?



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Re: M2102
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04 Oct 2014, 09:25
cjhande wrote: In option 2, The range (largest  Least )= 2b; how would be arrive at 2b from the set of the three numbers viz : a+b, ba, ab ? It's explained in the solution above: If a = b = 0, then S = {0, 0, 0} > the range = 0  0 = 0 = 2b. If a = 0 and b = 1, then S = {1, 1, 1} > the range = 1  (1) = 2 = 2b.
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Re: M2102
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25 Jul 2016, 05:32
a+b  (ab) = 2b, which is true for a=b=0 as well. Hence that makes ba as the median. From first option I got a+b as the median. What am I missing here?



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Re: M2102
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05 Sep 2016, 09:56
Option (1) states mean is a+b, which can be inferred from the question stem.
How does this lead a+b to 0? Could someone please clarify?



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Re: M2102
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05 Sep 2016, 22:25
manhasnoname wrote: Option (1) states mean is a+b, which can be inferred from the question stem.
How does this lead a+b to 0? Could someone please clarify? By simplifying this: \(mean=\frac{(a  b)+(b  a)+(a + b)}{3}=a+b\)
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Bunuel wrote: manhasnoname wrote: Option (1) states mean is a+b, which can be inferred from the question stem.
How does this lead a+b to 0? Could someone please clarify? By simplifying this: \(mean=\frac{(a  b)+(b  a)+(a + b)}{3}=a+b\) doesnt simlifying leads to: a+B/3=a+b ????????????



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Re: M2102
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31 Dec 2016, 13:49
I worked the second statement as following: Range= Highest Number  Lowest Number R=2b In order for R to be 2b, the only possible combination is (b+a)(ab). Thus, (ab)≤(ba)<(b+a) or (ab)<(ba)≤(b+a) So ba is the median. Where am I wrong?
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Judging by this question, you can't assume that the set provided to you is in ascending order? I'd feel more comfortable removing that assumption if anyone can provide an OG question which has a set of unknown variables not provided in ascending order.



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Re: M2102
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11 Feb 2017, 12:22
brooklyndude wrote: Judging by this question, you can't assume that the set provided to you is in ascending order? I'd feel more comfortable removing that assumption if anyone can provide an OG question which has a set of unknown variables not provided in ascending order. A set, by definition, is a collection of elements without any order. (While, a sequence, by definition, is an ordered list of terms.)
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Re: M2102
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17 Aug 2017, 08:37
Did not understand the inference of 1st option.
How the median of the set is a+b ??
looking at the set median is ba .



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Re M2102
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19 Aug 2017, 05:24
I think this is a highquality question and the explanation isn't clear enough, please elaborate. Hello expert, I get your explanation for option B, but aren't we worried about finding the mean in terms of a and b.
as explained even if a=b=0 or a=1, b=0, mean is ba.



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Re: M2102
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19 Aug 2017, 05:37
nitin083 wrote: I think this is a highquality question and the explanation isn't clear enough, please elaborate. Hello expert, I get your explanation for option B, but aren't we worried about finding the mean in terms of a and b.
as explained even if a=b=0 or a=1, b=0, mean is ba. The point is that in data sufficiency problems that ask for the value of a quantity, the data given in the statements are sufficient only when it is possible to determine exactly one numerical value for the quantity.
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Re: M2102
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06 Nov 2017, 05:47
+1 for A. My take : Statement 1 : Given that mean is a+b. (a+b)/3=a+b ; i.e a+b=0. This means that the set is 2b,0,2b. The median in this case becomes zero. We get a definite answer. Statement 2 : Range is 2b. The middle value is ba. This is not a definite value. Hence not sufficient. The answer is option A.
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Re: M2102
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06 Apr 2019, 21:58
Bunuel wrote: Official Solution:
(1) The mean of set \(S\) is equal to \(a + b\). Given that \(mean=\frac{(a  b)+(b  a)+(a + b)}{3}=a+b\), which leads to \(a+b=0\). Now, if \(a+b=0\), then \(ab\) and \(ba\) are either both zeros (if \(a=b=0\)) or have different signs (if \(a \ne b\)). In any case the median of \(S\) is \(a+b=0\). Sufficient. (2) The range of set \(S\) is equal to \(2b\). If \(a=b=0\), then \(median=0\) but if \(a=0\) and \(b=1\), then \(median=1\). Not sufficient.
Answer: A chetan2u,, BunuelWRT optionA, if A+B=0, then how is median =0. Do we assume that mean = median kindly explain Thanks



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Ashokshiva wrote: Bunuel wrote: Official Solution:
(1) The mean of set \(S\) is equal to \(a + b\). Given that \(mean=\frac{(a  b)+(b  a)+(a + b)}{3}=a+b\), which leads to \(a+b=0\). Now, if \(a+b=0\), then \(ab\) and \(ba\) are either both zeros (if \(a=b=0\)) or have different signs (if \(a \ne b\)). In any case the median of \(S\) is \(a+b=0\). Sufficient. (2) The range of set \(S\) is equal to \(2b\). If \(a=b=0\), then \(median=0\) but if \(a=0\) and \(b=1\), then \(median=1\). Not sufficient.
Answer: A chetan2u,, BunuelWRT optionA, if A+B=0, then how is median =0. Do we assume that mean = median kindly explain Thanks Got it overlooked the given set. Thanks










